Is angular velocity independent of initial force causing the motion?

Question

Context: I'm creating a simulated VR billiard game and trying to get the physics as realistic as possible. I have a cue ball being hit with a cue across a surface with accurate friction values calculated from real-world trials.

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Question: According to this source:

Angular velocity equals of a moving ball: $$w = \frac{5g}{2r}{ μ_k t}$$ If we look at the rolling phase after sliding, does this mean that no matter how hard you hit the ball, the angular velocity will always be the same (assume force acts on the center) as it is dependent on the radius and gravity only?

_{Hierrezuelo, J. and Carnero, C., 1995. Sliding and rolling: the physics of a rolling ball. Physics Education, 30(3), p.177.}

Answer 1

While a cue could be used to give the sphere some amount of initial angular momentum based on where it strikes, in this calculation it is assumed the sphere begins with zero rotation.

Since it begins with zero rotation, the angular acceleration depends only on the amount of friction, and the resistance to rotation due to the moment of inertia.

Now you asked if the velocity is always the same. What this equation says is the acceleration is always the same. But when the rotation matches the forward velocity, the acceleration stops. So the final velocity does depend on the initial energy/forward velocity imparted to the ball.

So if I hit a sphere exactly in the center pushing it forward with a force of 10N, would the angular acceleration be the same if I hit the ball with 20N?

Forces are applied over time. That's not what's happening here. The equation is what happens after the hit is done. In fact in the article, they separated it by assuming the ball was brought to some velocity and then placed on the table. We can't usefully talk about forces without considering the time of interaction, and we want that to be tiny. So better to talk about an impulse or short-duration change of momentum.

Assuming the hit is short enough, then yes, the forces don't matter. At the end of the hit, we assume the ball is not yet rotating, and moving with some forward velocity $v$. The greater the impulse, the greater the velocity.

After the hit is over, we assume friction with the table starts the ball rotating (and reducing the forward velocity). The rate this happens does not depend on the speed (in other words, the initial impulse), but at a constant depending on the friction and the ball.

When sufficient energy has moved from forward velocity to angular velocity, the ball stops skidding and the velocities no longer change in the ideal case. (Although they will continue to slow down due to drag in the real case).

Answer 2

I believe the problem is assuming you release the object from a constant initial velocity $v_0$ and the only forces acting on it are gravity, friction, and normal force. You are not applying an initial force $F_0$ or initial torque $\tau_0$ to the ball at the beginning by hitting it. If you did the physics would change.

The angular velocity $\omega$ is a function of time $t$ so it's changing over time. It is not constant. The angular acceleration $ \alpha = \frac{5g}{2r} \mu_k$ is what's constant. This makes sense cause you would expect a constant torque (in this case caused by the friction force $f$) to give a constant angular acceleration.